Question

A weak acid (HA) has a pKa of 4.895. If a solution of this acid has a pH of 4.516, what percentage of the acid is not ionized? (Assume all H in solution came from the ionization of HA.)

Answer 1

pH = 4.516

[H+] = 10^-pH = 10^4.516

[H+] = 3.05 x 10^-5 M

pKa = 4.895

Ka = 10^-pKa

Ka = 1.27 x 10^-3

HA -----------------> H+ + A-

C                           0           0 ---------------------> initial

C-x                        x           x -------------------------> after some time

Ka = (x^2) / C- x

but x = [H+] = 3.05 x 10^-5 M

1.27 x 10^-3 = (3.05 x 10^-5)^2 / (C - 3.05 x 10^-5)

C = 3.12 x 10^-5 M

percent of ionisaltion = (x / C ) x 100 = (3.05 x 10^-5) / 3.12 x 10^-5) x 100= 97.75%

percentage of the acid is not ionized = 100 -97.75 = 2.25 %